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254       XIV   INTEGRATION IN LOCALLY COMPACT GROUPS

1,^2)^1(^1)

f      r

=    d/22(x2)

J                  J

=

f(xl9

(*i> x2) ^i(X
by virtue of (13.21.2). Hence the result.

In particular, Lebesgue measure on R" (13.21.19) is a Haar measure on the
additive group R".

PROBLEMS

1.   Let G be a locally compact group, p, a left Haar measure on G, A a subset of G,
and B a relatively compact /z-integrable subset of G such that ^u(B) > 0. Show that, if
ft*(AB) < -f oo, then A is relatively compact (imitate the proof of (14.2.3)).

2.    (a)   Let G be a discrete subgroup of rank n in the additive group R", acting on R" by
translations. Show that, if A is Lebesgue measure, the number A(G) (Section 14.1,
Problem 6(b)) is equal to the absolute value of the determinant (with respect to the
canonical basis of R") of a Z-basis of G (use Problem 6(a) of Section 14.1). Deduce that
if A is a closed symmetric convex set in R" with nonempty interior (Section 12.14,
Problem 11) such that A(A) ^ 2"A(G), then A n G contains a point other than 0
(Minkowski's theorem).

n

(b)    Let Ui : (jc/)f  ^ CijXj (1  i ^ m) be linear forms on RM with integer coefficients

j=i

, and suppose that m < n. Let p be an integer > 1, and let A be a symmetric convex
set in R" with nonempty interior. Show that, for each r > 0 satisfying A(A)r" ^ 2npm,
there exists a point x ^ 0 in rA with integer coefficients, such that ut(x) = 0 (mod p)
for 1 < / < m. (Apply Minkowski's theorem to the subgroup GO of Z" consisting of all
z e Z" such that ut(z) ^ 0 (mod p) for 1 <; / g w, and use Problem 6(d) of Section 1 4.1 .)
In particular, show that if c1? c2 are any two integers, there exist integers xt, x2, not
both zero, such that |*i| ^Vp, \x2\ ^^/p and c^j. -f c2x2 =0 (mod p) (Thue's
theorem).

(c)    Let a, b be two integers. Use (b) to show that there exist integers xit x2 , #3, ^4,
not all zero, such that

55 x$ (mod />),  bxi  ax2  x4 (mod p)
and

y = xl -f xl + xl -f- xl

Show that if p is prime one can find two integers a, b such that a2 + b2 -f 1 = 0
(mod/?) (assuming/? is odd, observe that when z takes the J(P+ 1) integer valuesces /*, on sV for all s e G. Clearly a is left-